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[编程语言]CodeforcesRound#346(Div.2)


A. Round House
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Vasya lives in a round building, whose entrances are numbered sequentially by integers from 1 to n. Entrance n and entrance 1 are adjacent.
Today Vasya got bored and decided to take a walk in the yard. Vasya lives in entrance a and he decided that during his walk he will move around the house b entrances in the direction of increasing numbers (in this order entrance n should be followed by entrance 1). The negative value of b corresponds to moving |b| entrances in the order of decreasing numbers (in this order entrance 1 is followed by entrance n). If b?=?0, then Vasya prefers to walk beside his entrance.
[img]http://codeforces.com/predownloaded/e6/5a/e65af471e97813f0c1cba6e28d4be26b9af33a11.png Illustration for n?=?6, a?=?2, b?=??-?5.
Help Vasya to determine the number of the entrance, near which he will be at the end of his walk.
Input
The single line of the input contains three space-separated integers n, a and b (1?≤?n?≤?100,?1?≤?a?≤?n,??-?100?≤?b?≤?100) — the number of entrances at Vasya's place, the number of his entrance and the length of his walk, respectively.
Output
Print a single integer k (1?≤?k?≤?n) — the number of the entrance where Vasya will be at the end of his walk.
Examples
Input
6 2 -5

Output
3

Input
5 1 3

Output
4

Input
3 2 7

Output
3

Note
The first example is illustrated by the picture in the statements.
题意:1~n的环,从a走b步,会走到那个位置;b>0表示顺时针走,b<0表示逆时针走。
分析:水题~~


<span style="font-size:18px;">#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
#define ll long long
#define CL(a,b) memset(a,b,sizeof(a))
#define lson (i<<1)
#define rson ((i<<1)|1)
#define MAXN 100010

int main()
{
    int n,a,b;
    while(cin>>n>>a>>b)
    {
        int t;
        if(b > 0)
            t = (a+b%n)%n;
        else
            t = a-(-b)%n;
        if(t <= 0)t=n+t;
        printf("%d\n",t);
    }
    return 0;
}
</span>


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2016-04-03 20:43:25  
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